c_1 +2 C_2 C_1 +3 C_3 C_2 + +n C_n C_ n-1 =? - Vidyadip

MCQ

$\left\{\frac{c_1}{\infty}+2 \frac{C_2}{C_1}+3 \frac{C_3}{C_2}+\ldots+n \cdot \frac{C_n}{C_{n-1}}\right\}=?$

A
$\frac{1}{2} n(n+1)$
B
2n
C
$2^{n-1}$
D
$2^n$

✓

Answer

(a) $\frac{1}{2} n(n+1)$
Explanation: We know that $\frac{C_r}{C_{r 1}}=\frac{n-r+1}{r}$,
Substituting $r =1,2,3, \ldots, n$, we obtain
$\frac{C_1}{C_0}+2 \cdot \frac{C_2}{C_1}+3 \cdot \frac{C_3}{C_2}+\ldots+n \cdot \frac{C_n}{C_{n-1}}= n +( n -1)+( n -2)+\ldots+1=\frac{1}{2} n(n+1)$.

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